Mrs. McIntyre needs to buy pencils and erasers for her math class. The store is selling a box of pencils for $2 and a box of era
1 answer:
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Answer:
1) x ≤ 2 or x ≥ 5
2) -6 < x < 2
Step-by-step explanation:
1) We have x^2 - 7x + 10, so let's factor this as if this were a regular equation:
x^2 - 7x + 10 = (x - 2)(x - 5)
So, we now have (x - 2)(x - 5) ≥ 0
Let's imagine this as a graph (see attachment). Notice that the only place that is above the number line is considered greater than 0, and that's when x ≤ 2 or x ≥ 5 (the shaded region).
2) Again, we have x^2 + 4x - 12, so factor this as if this were a regular equation:
x^2 + 4x - 12 = (x + 6)(x - 2)
So now we have (x + 6)(x - 2) < 0
Now imagine this as a graph again (see second attachment). Notice that the only place that is below 0 (< 0) is when -6 < x < 2 (the shaded region).
Hope this helps!
Answer:
The missing figure is attached down
The length of BE is 27 units ⇒ 3rd answer
Step-by-step explanation:
In circle A:
- ∠BAE ≅ ∠DAE
- Line segments A B, A E, and A D are radii
- Lines are drawn from point B to point E and from point E to point D to form secants B E and E D
- The length of B E is 3 x minus 24 and the length of E D is x + 10
We need to find the length of BE
∵ AB and AD are radii in circle A
∴ AB ≅ AD
In Δs EAB and EAD
∵ ∠BAE ≅ ∠DAE ⇒ given
∵ AB = AD ⇒ proved
∵ EA = EA ⇒ common side in the two triangles
- Two triangles have two corresponding sides equal and the
including angles between them are equal, then the two
triangles are congruent by SAS postulate of congruence
∴ Δ EAB ≅ Δ EAD ⇒ SAS postulate of congruence
By using the result of congruence
∴ EB ≅ ED
∵ EB = 3 x - 24
∵ ED = x + 10
- Equate the two expressions to find x
∴ 3 x - 24 = x + 10
- Add 24 to both sides
∴ 3 x = x + 34
- Subtract x from both sides
∴ 2 x = 34
- Divide both sides by 2
∴ x = 17
Substitute the value of x in the expression of the length of BE to find its length
∵ BE = 3 x - 24
∵ x = 17
∴ BE = 3(17) - 24
∴ BE = 51 - 24
∴ BE = 27
The length of BE is 27 units
